 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Derivatives of Power Functions - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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To find the equation of a tangent line, first sketch the tangent line. Remember that a tangent line touches the curve of a function at only one point. After sketching the tangent line, find any two points (x1, y1) and (x2, y2) on it. From these two points, you can calculate the slope m (with the usual formula of the change in y divided by the change in x). Then, find the y-intercept of the graph by plugging these points into point-slope form: y-y1=m(x-x1). This will give you the equation of the tangent line. When you have the equation of the tangent line, you can calculate the derivative of the graph at that point x by plugging in the value of the x you used to draw the tangent line.

Let's take a look at another problem. Suppose I have the function f(x) equals 1 over x² which I have graphed here. First, let's sketch the line tangent to f at x equals -2.

So let me pick another color. Here is x equals -2 right here. I want to sketch the tangent line. One thing that's good about sketching tangent lines ahead of time, is in part b I'm going to find an equation for this line, and having a sketch that will help me make sure that I get the right answer. Looking at this sketch, I can see that I expect a positive slope.

Let's find an equation for the tangent line that I just graphed. So f(x) is 1 over x². That's the same as x to the -2. So the derivative of this f' is -2x to the -3 which is the same as -2 over x³.

Now, to find the equation of tangent line I need two things.; a point, and a slope. For the point at x equals -2, I need a y coordinate f(-2), so that's 1 over -2² or ¼. So my point is -2,¼.

Now what's the slope? The derivative gives me slope. So f'(-2) is -2 over -2³. So that's 2 over -8 which is also ¼, so the slope is ¼.

So in point-slope form, my tangent line is going to be y minus the y coordinate, ¼, equals the slope which is also ¼, times x minus the x coordinate. So x plus 2. So this is an equation of tangent line in Point-slope form.

If you want it slope-intercept form, all you have to do is solve this for y. So you have ¼x. ¼ times 2 is 1/2, and I add ¼ to that and I get 3/4. So y equals 1/4 x plus 3/4.

Let me go back to the graph. That's the equation of line line. 1/4x plus 3/4. The equation of the tangent to f at x equals -2.